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Reciprocal of a vector

  1. Dec 17, 2013 #1
    [tex]\vec{v}=\frac{\mathrm{d} \vec{r}}{\mathrm{d} t}[/tex]
    [tex]\\ \vec{v} = \frac{d\vec{r}}{dt} \\ \\ \vec{v}\;dt = d\vec{r} \\ \\ dt = \frac{d\vec{r}}{\vec{v}} \\ \\ \int dt = \int \frac{d\vec{r}}{\vec{v}} \\ \\ t = \int \frac{d\vec{r}}{\vec{v}}[/tex]
    Is true?
    Solving the equation for time t, is need divide the position vector r by velocicty vector v... But I don't know do division between vectors...
  2. jcsd
  3. Dec 17, 2013 #2

    D H

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    Division by a vector is not defined.

    It appears you are trying to solve some problem and you thought this might be a way to attack that problem. What is the problem that are you trying to solve?
  4. Dec 17, 2013 #3
    Given a vectorial velocity and a position vector, I want to calcule o time t.
    In other words, I want solving the equation above for variable t.
    [tex]\\ \vec{v} = \frac{d\vec{r}}{dt}[/tex]
  5. Dec 17, 2013 #4


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    In general, if you are given [itex]\vec{v}[/itex] and [itex]\frac{d\vec{r}}{dt}[/itex], there will NOT BE a variable t such that [tex]\vec{v}= \frac{d\vec{r}}{dt}[/tex]!

    You can try [tex]\int \vec{v}dt= \int\vec{dr}[/tex] and then solve the resulting equation for t.
  6. Dec 17, 2013 #5


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    [itex] \vec{r} \vec{u}= \vec{r} \frac{d\vec{r}}{dt}[/itex]
    [itex]\vec{r} \vec{u} =\frac{1}{2} \frac{dr^{2}}{dt}[/itex]
    [itex]dt= \frac{dr^{2}}{\vec{r} \vec{u}}=\frac{dr^{2}}{ru cosθ}[/itex]

    is that correct?
  7. Dec 18, 2013 #6
    In this case, the only solution is to consider the modulus of vectos: t = ∫ 1/v dr. Correct!?
  8. Jan 5, 2014 #7
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